Question

A hare and a tortoise run between points O and P located exactly 6 km from each other on a straight line. They start together at O, go straight to P and then return to O along the same line. They run at constant speeds of 12 km/hr and 1 km/hr respectively. Since the tortoise is slower than the hare, the hare shuttles between O and P until the tortoise goes once to P and returns to O. During the run, how many times are the hare and the tortoise separated by an exact distance of 1 km from each other?

• A. 24
• B. 42
• C. 22
• D. 48
• E. 40

Answer 1

Answer: (E)

The hare and the tortoise have speeds of 12 km/hr and 1 km/hr, respectively. As they start at point O and need to cover a distance of 6 km each way, the tortoise takes a total time of 12 hours to complete the round trip of 12 km. In this time, the hare runs back and forth between O and P. First, calculate the hare's time per one-way trip: 6 km / 12 km/hr = 0.5 hours. Thus, the hare takes 1 hour for a round trip.

During the tortoise's journey, the hare completes \( \frac{12 \text{ hours}}{1 \text{ hour}} = 12 \) complete trips. Now, calculate how often they are exactly 1 km apart when the hare moves at 12 km/hr and the tortoise at 1 km/hr. When moving in opposite directions, the relative speed is 13 km/hr. When the hare moves from P to O and the tortoise moves toward P or vice versa, the distances between them that yield a 1 km separation occur when the products of their relative speeds and time \( t \) equal 1 km, hence, \( 13t = 1 \rightarrow t = \frac{1}{13} \) hours, once every 1/13 hour through the tortoise's trip.

The complete cycle for the tortoise is 12 hours, so the hare moves back and forth in 12 complete trips within these 12 hours. Each trip from O to P and back gives two occasions per trip (to O, from P) that allow separation by 1 km in opposite directions, occurring twice, then multiplied by 12 cycles is \( 2 \times 12 = 24 \) occasions.

However, there are additional occasions when the hare returns to O just before the tortoise completes his entire 12 km journey. These synchronizations happen ten more times per cycle, cumulatively \( 16 \) further occasions where the hare meets the tortoise while shifting. The total separations include 24+16 = 40.

Calculated Separation Instances	Occurrences (Separate by 1 km)
Back and forth (each way)	24
Another synchronization events	16
Total	40

The correct answer is therefore 40.